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On $E=\mathbb{R}$, take $f_n(x)=x$ and $g_n(x)=1/n$. Then $f_n\to f(x)=x$ with uniform convergence, because there is no error, and $g_n\to 0$ with uniform convergence, with error $1/n$. Their products $f_n{g}_n=x/n$ converge pointwise to zero, but for every $n$ the choice $x=n$ gives $|f_n(x)g_n(x)|=1$. Thus the products do not converge uniformly. Here the first family consists of unbounded functions, exactly the missing hypothesis in the preceding estimate.

Write $f_n\to f$ and $g_n\to g$ with uniform convergence. The triangle inequality givesHence $f_n+g_n\to f+g$ with uniform convergence.

If every $f_n,g_n$ is a bounded function, part iii gives constants $M_f,M_g$ bounding the two families. Their pointwise limits satisfy $|f|\le M_f$ and $|g|\le M_g$. ConsequentlyThus both sums and products have uniform convergence under the stated hypotheses. The product estimate is the uniform convergence product rule for bounded functions.

Assume, as required for the displayed integrals, that the $f_n$ are Riemann integrable on $[a,b]$. Their uniform limit $f$ is also Riemann integrable: if $\sup |f-f_N|<\epsilon$, the upper and lower Darboux sums for a partition of an interval differ for $f$ by at most the corresponding difference for $f_N$ plus $2\epsilon (b-a)$. Choosing $N$ and then a sufficiently fine partition of an interval proves the Riemann integrability criterion for $f$.

The integral estimate then givesThis proves that uniform convergence preserves integrals on a compact interval. Some integrability hypothesis is necessary: uniform convergence alone cannot make the integral of an arbitrary starting function exist.

No: the unbounded length of the domain matters. Let $\varphi (t)=\max \{1-|t|,0\}$ and define $f_n(x)=n^{-1}\varphi (x/n)$ on $\mathbb{R}$. These are continuous functions with compact support, and ${\sup }_{\mathbb{R}}|f_n|=1/n\to 0$, so the sequence has uniform convergence to zero. However, the change of variables $t=x/n$ givesThe integral of the limit is zero. The supports widen as the heights decrease; a uniform bound on the heights gives no bound on the total area over an infinite interval.

Take $f_n(x)=x^n$ on $[0,1]$. Each function is continuous, and the pointwise limit is $f(x)=0$ for $0\le x<1$, while $f(1)=1$. NeverthelessThe limiting function is Riemann integrable, since changing a value at one point does not change its Riemann integral. There is no uniform convergence: for every $n$, ${\sup }_{[0,1]}|f_n-f|=1$, with the supremum approached as $x\uparrow 1$ from below. Equivalently, the uniform limit theorem would require the limit to be continuous.

The truncated Fourier cosine series gives the better approximation over the whole closed interval. Its Fourier coefficients obey $|a_n|\le 4/(\pi n^2)$, so absolute convergence and the tail boundgive uniform convergence. The even periodic extension is continuous; its slope changes only cause corners.

By contrast $b_n=(-1{)}^{n+1}/n+O(n^{-2})$. The odd periodic extension has a jump at $\pi$, giving slower coefficient decay and the Gibbs phenomenon near that endpoint. Every truncated Fourier sine series satisfies $S_N(\pi )=0$, soIt therefore cannot converge uniformly to $f$ on $[0,\pi ]$, although it converges at every point of $[0,\pi )$. This comparison concerns approximation across the interval; it does not assert that the cosine truncation has smaller error at every individual interior point for every $N$.

Write $T=T_s+T_u$. The transient satisfies the heat equation and homogeneous mixed boundary conditionsThe left condition is a Neumann boundary condition and the right one a Dirichlet boundary condition. Separation of variables, $T_u=X(x)G(t)$, givesThe condition $X^{\prime }(0)=0$ removes the sine term. The condition $X(2\ell )=0$ then selectsThere is no nonzero mode with $\mu =0$, because a linear $X$ satisfying both conditions is zero. There is no growing mode either: integration by parts gives ${\int }_0^{2\ell }(X^{\prime }{)}^{2}dx={\mu }^2{\int }_0^{2\ell }{X}^{2}dx$, so every nonzero mode has ${\mu }^2>0$.

These Sturm-Liouville eigenfunctions have orthogonality and squared normExpand the initial transient in this half-integer Fourier cosine series. Its coefficients areThe integration formula follows by integration by parts; the cosine at the endpoint is zero. Thus the half-integer cosine expansion of a linear profile givesAt $t=0$ the Fourier series reproduces $-T_s$ everywhere, including the endpoints; its $n^{-2}$ coefficient decay gives uniform convergence. For every $t>0$ the exponential factors allow termwise differentiation and verify the heat equation and both boundary conditions. The sudden switch-on of nonzero heat flux is incompatible with the spatial derivative of the constant initial temperature at the corner $(0,0)$. Accordingly, the positive-time boundary derivative must not be inferred by differentiating the initial Fourier series term by term at that corner.

On the basic interval the graph is the pair of straight segments joining $(-\pi ,0)$, $(0,1)$ and $(\pi ,0)$. Its periodic extension is an even triangular wave.

The Fourier coefficients satisfy $b_n=0$ by evenness, andFor the second formula, integration by parts gives ${\int }_0^{\pi }x\cos (nx)dx=[(-1{)}^n-1]/n^2$. Only odd indices survive, giving this rescaled Fourier cosine series of a triangular wave:The pointwise convergence of a piecewise smooth Fourier series applies throughout the interval, including the continuous joined endpoints. The coefficients also imply uniform convergence by the Weierstrass M-test. At $x=0$, the value is one, so

The periodically extended $x^2$ is even, so every sine coefficient is zero. With the convention $f=a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$, the Fourier coefficients areFor the second integral, two applications of integration by parts give ${\int }_0^{\pi }{x}^2\cos nxdx=2\pi (-1{)}^n/n^2$. ConsequentlyThe periodic extension is continuous even at the joined endpoints, where both one-sided values equal ${\pi }^2$. The pointwise convergence of a piecewise smooth Fourier series therefore gives the displayed values throughout the interval; the coefficient bound also gives uniform convergence by the Weierstrass M-test. Setting $x=\pi$ and then $x=0$ yields

Evenness removes every sine coefficient. In the period-two Fourier cosine series convention $f=a_0/2+{\sum }_{n\ge 1}{a}_n\cos (n\pi x)$, the constant coefficient is $a_0=2{\int }_0^1(1-x)dx=1$. Integration by parts givesThus the Fourier cosine series of a triangular wave isThe periodic function is continuous and piecewise smooth, so the Dirichlet-Jordan convergence theorem gives its value at every point. The summable absolute coefficient bound also gives uniform convergence by the Weierstrass M-test.

Use the real Fourier series convention $f=a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$. The function is an even function, so $b_n=0$. For noninteger real $\mu$,For $n\ge 1$, the product-to-sum identity givesThus the Fourier series of a noninteger-frequency cosine isIts Fourier coefficients are $O(n^{-2})$, giving absolute convergence and uniform convergence. The periodic function is continuous at its joins because the two endpoint cosine values coincide, so the Dirichlet-Jordan convergence theorem gives this value at every point.

Putting $\mu =1/3$ givesEvaluate at $x=0$, where $f(0)=1$, and multiply by $\pi /\sqrt{3}$:Comparing with the required form gives $\boxed{A=3,B=3}$.

The power series for $e^{y^3}$ has uniform convergence on every compact interval, so termwise integration from zero to $x$ is valid. ThusThis power series converges for every real $x$, including negative $x$ with the usual orientation of a definite integral.

Steady state. In the heat equation, time independence makes the second spatial derivative zero. The two endpoint temperatures therefore determine the linear profileInsulation is along the sides; the endpoint reservoirs can still exchange heat with the rod.

Transient solution. Subtract the steady profile before applying separation of variables: let $U=T-T_{0}x/l$. Then $U_t=\nu U_{xx}$, $U(0,t)=U(l,t)=0$ and $U(x,0)=-T_{0}x/l$ for $0<x<l$. Writing $U=X(x)\Theta (t)$ givesThe homogeneous Dirichlet boundary conditions exclude $\mu \le 0$ and select the eigenfunctions $\sin (n\pi x/l)$ with eigenvalues ${\mu }_n=(n\pi /l{)}^2$, $n\ge 1$. Their orthogonality gives the Fourier sine series coefficientsHere integration by parts uses ${\int }_0^{l}x\sin (n\pi x/l)dx=-l^2(-1{)}^n/(n\pi )$. ThusEvery summand satisfies the heat equation and the homogeneous endpoint conditions, while the steady profile supplies the prescribed endpoint values. The initial Fourier sine series equals $-T_{0}x/l$ in the open interval. At $(l,0)$ the initially cold rod and the suddenly imposed hot endpoint are incompatible; the formula describes the prescribed data for $t>0$ and the initial limit at interior points.

Heat absorption. Integrating in $x$ kills the even modes. With $a=\nu {\pi }^2/l^2$, the total heat relative to the initial state isConsequently the net rate of heat absorption isFor each $t\ge \epsilon >0$, any fixed number of differentiated modes is bounded by a polynomial in $n$ times $e^{-a\epsilon n^2}$. These bounds have finite sums, proving absolute convergence, uniform convergence on such time intervals and the validity of the integration and differentiation. In particular the displayed rate is finite for every $t>0$, positive when $T_0>0$, and tends to zero as equilibrium is approached. The same answer follows from the heat equation balance $H^{\prime }=\gamma \nu [T_x(l,t)-T_x(0,t)]$; this subtracts loss at the cold endpoint from input at the hot endpoint.

Multiplication by $x^2$ puts the eigenvalue problem into Sturm-Liouville form, $-(x^2{y}^{\prime }{)}^{\prime }=\lambda x^{2}y$, with weight $x^2$. If $u,v$ correspond to distinct eigenvalues $\lambda ,\mu$, subtract their multiplied equations and integrate to getThe upper boundary term vanishes by the Dirichlet boundary condition. For a bounded solution near zero, $x^2{y}^{\prime }$ has a limit because $(x^2{y}^{\prime }{)}^{\prime }=-\lambda x^{2}y$ is integrable. A nonzero limit would force a $1/x$ divergence in $y$, so that limit is zero; integration then gives $y^{\prime }=O(x)$. The lower boundary term therefore also vanishes, proving the weighted orthogonality.

Set $z=xy$. For $x>0$, $y^{\prime \prime }+2y^{\prime }/x=z^{\prime \prime }/x$, and finiteness of $y(0)$ gives $z(0)=0$. Thus $z^{\prime \prime }+\lambda z=0$ with zero values at zero and $\pi$. The radial Dirichlet spectrum in three dimensions givesThe zero and negative eigenvalues yield only the zero solution under these boundary conditions. Also ${\int }_0^{\pi }{x}^2{y}_n^{2}dx=\pi /2$.

For the forced equation, the same substitution gives $z^{\prime \prime }-\beta z=x$. The Fourier sine series of $x$ on $(0,\pi )$ has coefficientsExpanding $z=\sum c_n\sin nx$ gives $-(n^2+\beta )c_n=b_n$, henceThe value at zero is defined by a limit. In fact $|\sin nx/x|\le n$, so uniform convergence holds even there, by comparison with $2\sum (n^2+\beta {)}^{-1}$. An independent check comes from solving the constant-coefficient equation directly:It satisfies both endpoint conditions and the forcing for $x>0$, and its finite origin limit agrees with the series. For $\beta >0$ the homogeneous boundary value problem has no nonzero solution, so the solution is unique.

Set $T(x,t)=X(x)G(t)$ in the heat equation. Separation of variables gives $G^{\prime }/(c^{2}G)=X^{\prime \prime }/X=-\lambda$. Zero endpoint temperatures impose Dirichlet boundary conditions $X(0)=X(\pi )=0$, selecting $\lambda =n^2$ for $n\ge 1$. The separated factors can therefore be chosen asZero and negative separation eigenvalues give only the zero spatial solution under these boundary conditions.

The initial profile is continuous and triangular, with midpoint height $\pi /2$. Its Fourier sine series coefficients areConsequently the heat flow from a triangular initial temperature isFor $t\ge 0$, each summand has magnitude at most $4/[\pi (2m+1{)}^2]$. The Weierstrass M-test proves absolute and uniform convergence. The initial Fourier sine series equals the continuous piecewise smooth profile, and for $t>0$ the exponential factors justify termwise spatial and time differentiation on every time interval bounded away from zero.

Write $\rho =|\mathbf{x}-\mathbf{y}|$. A rotationally symmetric fundamental solution of the Laplace equation is harmonic for $\rho >0$. The Laplace equation in polar coordinates reduces to $(\rho G^{\prime }{)}^{\prime }=0$, so $G=A\log \rho +B$. Integrate $-\Delta G={\delta }^{(2)}(\mathbf{x}-\mathbf{y})$ over a disk of radius $ϵ$ around the source. The divergence theorem givesTaking $B=0$ therefore givesChanging the reference length inside the logarithm merely adds a constant. Notice the sign convention: the potential equation in this question is $\Delta V=q{\delta }^{(2)}$, whereas this Green function satisfies $-\Delta G={\delta }^{(2)}$. Thus the point-source potential used below is $-qG=+(q/2\pi )\log \rho$.

For the two insulating walls, successive Euclidean reflections in $x=b$ and $x=-b$ generate sources of strength $q$ at $(2nb,0)$ for every integer $n$. Only $n=0$ lies inside the physical strip. This is the Neumann image lattice for a point source in a strip. SetThe raw sum of logarithmic potentials diverges by additive constants. A well-defined potential is obtained by pairing opposite images and subtracting their constant reference terms:Each paired summand is $O(n^{-2})$ on compact subsets away from the source. The series therefore has local uniform convergence, as do the required derivative series. It givesIn the corresponding $x$-derivative, the images pair as $n$ and $1-n$ at $x=b$, and as $n$ and $-1-n$ at $x=-b$. Their opposite horizontal separations make the derivatives cancel. Thus both homogeneous Neumann boundary conditions hold. The constructed solution is even in $y$; without a condition at infinity the side conditions alone would allow further homogeneous harmonic functions.

For fixed $y>0$, uniform convergence on $[-b,b]$ permits integration term by term. The $n$th term has integralHere $\arctan$ is the ordinary real inverse trigonometric function, taking values in $(-\pi /2,\pi /2)$. Summing from $-N$ to $N$ makes a telescoping series:The source flux splits equally between the upper and lower parts of the strip. The next two parts give the single-wall constructions that underlie the image lattice.

For the usual piecewise smooth complex path $\gamma$, let $\ell (\gamma )$ denote its finite length. The estimation lemma givesBy uniform convergence, the supremum tends to zero, soThe uniform limit is continuous on the path, so its integral exists. This is uniform convergence and contour integration; the path length is fixed while the functions vary.

Fix any bounded disk avoiding the integers, and choose $R$ with $|z|\le R$ on it. For sufficiently large $|n|$, say $|n|\ge 2R$ and $|n|\ge 1$,The convergent numerical series ${\sum }_{n\ne 0}{n}^{-2}$ and the Weierstrass M-test give absolute and uniform convergence of the tails. The finitely many remaining summands are holomorphic away from their integer poles. Thus the partial sums converge locally uniformly on $\mathbb{C}\setminus \mathbb{Z}$.

By part (a)(iii), the sum is a holomorphic function on $\mathbb{C}\setminus \mathbb{Z}$. The term $n=0$ is $-1/z^2$ and is harmless on this domain; it must not be omitted.

First verify that $V_0$ is a vector space. Sums and scalar multiples preserve continuity and the condition at zero. For every partition of an interval $D$, the triangle inequality and absolute homogeneity of a norm of the absolute value implyTaking suprema yieldsFor $\lambda =0$ the second equality follows directly from $V(0)=0$. In particular, finite variation is preserved, so $V_0$ is closed under these operations.

Nonnegativity is immediate. For any $t\in [0,1]$, use the partition of an interval $(0,t,1)$, omitting repeated endpoints if necessary, to getThus $V(f)=0$ forces $f=0$, and conversely $f=0$ has zero total variation of a function. These observations prove that $\Vert f\Vert =V(f)$ is a norm on $V_0$.

Applying the same bound to $f_n-f$ givesTherefore convergence in the anchored variation norm implies uniform convergence on $[0,1]$. The anchoring condition $f(0)=0$ removes the nonzero constant functions which otherwise have zero total variation of a function.

Truncate the Chebyshev series at the full degree of precision of a quadrature rule:Since $|T_j(x)|\le 1$ on $[-1,1]$, the geometric series estimate givesThe same bound proves uniform convergence of the original Chebyshev series, so its sum and the tail estimate are justified. Gauss-Legendre quadrature is exact on $s_{2n+1}$. Its positive quadrature weights sum to $2$, henceTherefore one explicit choice proving exponential convergence of Gaussian quadrature isUsing exactness through polynomial degree $2n+1$ doubles the exponent compared with truncating at polynomial degree $n$.